0.00/0.00	NO
0.00/0.00	
0.00/0.00	
0.00/0.00	Succeeded in reading "/export/starexec/sandbox/benchmark/theBenchmark.trs".
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR z x)
0.00/0.00	    (RULES
0.00/0.00	      s(a) -> c
0.00/0.00	      s(b) -> c
0.00/0.00	      c -> t(k)
0.00/0.00	      c -> t(l)
0.00/0.00	      f(x) -> z | s(x) == z
0.00/0.00	      g(x,x) -> h(x,x)
0.00/0.00	    )
0.00/0.00	    (COMMENT [75] Example 41 submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	No "->="-rules.
0.00/0.00	
0.00/0.00	Decomposed conditions if possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR z x)
0.00/0.00	    (RULES
0.00/0.00	      s(a) -> c
0.00/0.00	      s(b) -> c
0.00/0.00	      c -> t(k)
0.00/0.00	      c -> t(l)
0.00/0.00	      f(x) -> z | s(x) == z
0.00/0.00	      g(x,x) -> h(x,x)
0.00/0.00	    )
0.00/0.00	    (COMMENT [75] Example 41 submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	Removed infeasible rules as much as possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR z x)
0.00/0.00	    (RULES
0.00/0.00	      s(a) -> c
0.00/0.00	      s(b) -> c
0.00/0.00	      c -> t(k)
0.00/0.00	      c -> t(l)
0.00/0.00	      f(x) -> z | s(x) == z
0.00/0.00	      g(x,x) -> h(x,x)
0.00/0.00	    )
0.00/0.00	    (COMMENT [75] Example 41 submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	Try to disprove confluence of the following (C)TRS:
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR z x)
0.00/0.00	    (RULES
0.00/0.00	      s(a) -> c
0.00/0.00	      s(b) -> c
0.00/0.00	      c -> t(k)
0.00/0.00	      c -> t(l)
0.00/0.00	      f(x) -> z | s(x) == z
0.00/0.00	      g(x,x) -> h(x,x)
0.00/0.00	    )
0.00/0.00	    (COMMENT [75] Example 41 submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	Succeeded in disproving confluence.
0.00/0.00	
0.00/0.00	Disproved via the following (C)TRS:
0.00/0.00	(CONDITIONTYPE ORIENTED)
0.00/0.00	(VAR z x)
0.00/0.00	(RULES
0.00/0.00	  s(a) -> c
0.00/0.00	  s(b) -> c
0.00/0.00	  c -> t(k)
0.00/0.00	  c -> t(l)
0.00/0.00	  f(x) -> z | s(x) == z
0.00/0.00	  g(x,x) -> h(x,x)
0.00/0.00	)
0.00/0.00	(COMMENT [75] Example 41 submitted by: Thomas Sternagel and Aart Middeldorp)
0.00/0.00	
0.00/0.00	NO
0.00/0.00	EOF